Find the value of a and b using complex numbers

Question

Here is a question:

$$1+3i$$

is a root of

$$z^2+az+b=0,$$ where a and b are real.

Find $a$ and $b$.

Answer 1

By the complex conjugate root theorem, the other root is $1-3i$. Therefore,

$z^2+az+b=(z-(1+3i))(z-(1-3i))$

$=z^2-(1+3i+1-3i)z+(1+3i)(1-3i)=z^2-2z+10.$

Can you take it from here?

Answer 2

-a is the sum of the roots and c is the product, so

$$a = -(1+3i) - (1-3i) = -2$$

and

$$b = (1+3i)(1-3i) = 1 + 9 = 10$$

Answer 3

If

$a, b \in \Bbb R, \tag 1$

and

$z \in \Bbb C \setminus \Bbb R \tag 2$

satisfies

$x^2 + ax + b \in \Bbb R[x]; \tag{2.5}$

then

$z^2 + az + b = 0; \tag 3$

taking the complex conjugate we find

$\bar z^2 + \bar a \bar z + \bar b = 0 \tag 4$

as well; by virtue of (1),

$\bar a = a, \; \bar b = b, \tag 5$

and thus (4) becomes

$\bar z^2 + a \bar z + b = 0; \tag 6$

from (2)

$z \ne \bar z, \tag 7$

so $z$ and $\bar z$ are the distinct roots of (2.5); thus,

$x^2 + ax + b = (x - z)(x - \bar z) = x^2 - (z + \bar z)x + z \bar z, \tag 8$

that is,

$a = -(z + \bar z), \; b = z \bar z; \tag 9$

if we take

$z = 1 + 3i, \; \bar z = 1 - 3i, \tag 8$

then

$a = -2, \; b = 10. \tag{9}$